Solve the equation. $\dfrac{dy}{dx}=-\dfrac{1}{3}x^3$ Choose 1 answer: Choose 1 answer: (Choice A) A $y=-\dfrac{x^4}{12}+C$ (Choice B) B $y=-x^2+C$ (Choice C) C $y=-\dfrac{x^4}{12+C}$ (Choice D) D $y=-x^{2+C}$
Explanation: We can bring this equation to the form $f(y)\,dy=g(x)\,dx$ : $\begin{aligned} \dfrac{dy}{dx}&=-\dfrac{1}{3}x^3 \\\\ -3\,dy&=x^3\,dx \end{aligned}$ This means we can solve this equation using separation of variables! $\begin{aligned} -3\,dy&=x^3\,dx \\\\ \int -3\,dy&=\int x^3\,dx \\\\ -3y&=\dfrac{x^4}{4}+C_1 \\\\ y&=-\dfrac{x^4}{12}+C \end{aligned}$ [Where did we get C?] Notice that after the integration, more work was required in order to isolate $y$. In conclusion, this is the solution of the equation: $y=-\dfrac{x^4}{12}+C$